# Question on moment of inertia

Wen
I have been thinking for an hour and i still couldn't get the answer for this qns. It involve moment of inertia.

a uniform solid disk, radius R and mass M, free to rotate about a frictionless pivot on a point on its rim.It is released from a position where the centre of mass is horizontal to its pivot, and its allowed to swing till the centre of mass is vertically below. What is the speed of the centre of mass when is at that position?

My solution

moment of inertia=Icm+MR^2
moment of inertia= 1/2 MR^2 +MR^2=3/2 MR^2
Torque=moment of inertia X angular a
Mg R= 3/2 MR^2 a
a=2g/3R
Final angular v (Wf)^2= Wi^2 +2a(angular displacement)
Wf^2= 0 +2(2g/3R)(pi/4)
Wf can be found
V of the centre of mass , R distance away from the pivot= RWf

However, the answer in the book is 2 (Rg/3)^0.5

## Answers and Replies

Homework Helper
Think ... Whether the torque is constant?(NOOO). The angular acceleration is non uniform. The equation used, is it true for non uniform accelerations?

Batter to use energy conservation.

Homework Helper
The torque is variable.
Instead of being T = MgR, it's more like T = MgR.cosθ

Hint: d²θ/dt² = dω/dt = dω/dθ.dθ/dt = ω.dω/dθ

so use: d²θ/dt²= ω.dω/dθ

Wen
So i get MgRcosQ=3/2R.w.dw/dQ

I am stuck again?

Wen
Hurray!!!
I got it!!!
MgRcos Q=I A Q=angle btw g and the F(in the direction of v)
I cm = integrate r^2dm(limit:R to 0)

=inte.r^2(2pir)DXdr X=thickness, D=density
....
=1/2MR^2
Since axis is not abt the COM

I= Icm+MR^2
=3/2MR^2

.;MgcosQ=3/2MR^2.A
A can be found

Wf^2=Wi^2+2AQ
=0+2{2gCosQ/3R)dQ
Since Q varies from 0 to the position( pi/4)

Wf^2=inte. 4gcos Q/3R. dQ (limit: pi/4 to 0)
=[4gSinQ/3R]
=4g/3R
Wf =root 4g/3R
V =R root 4g/3R

Thanks everyone

Homework Helper
Can be done using energy conservation as

loss in PE = gain in RKE
MgR = 0.5 I w^2
MgR = 0.5(1.5MR^2)v^2/R^2
v^2 = 4gR/3

j3w3ls
What if, instead of uniform solid disk, a uniform hoop is used? what's the Vcm then? Thanks!